Introduction¶

Learning outcomes¶

Understand hydrostatic forces on inclined surfaces
Learn about and how to use the concept of the pressure prism
Understand hydrostatic forces on curved surfaces
Learn about buoyancy and how heavy things like ships float

Hydrostatic force on plane surfaces¶

Consider a large rectangular water tank such as a swimming pool. We've already looked at the effect of depth on pressure but now we want to look as the forces exerted by the fluid on its surroundings.

We want to understand the fores the volume of fluid exerts on the bottom and side walls.

Which we can simplify this example into a 2D problem as shown below. Note that the inverted triangle represents the level of the free surface which is subject to atmospheric pressure. We can consider it analogous to an electrical ground.

For the forces acting on the bottom surface (area $A$) we can simply integrate over the pressure distribution on that surface.

\begin{equation} F_R = \int_A p dA = \gamma h A \end{equation}

However on the side walls the calculation is more complicated since the the pressure distribution on the side walls increases with depth. As shown above it is zero at the surface and increases to some maximum value. This is an important engineering problem as it determines the structural strength required in many engineering areas such as hydroelectric power generation, marine engineering and civil engineering.

Hydrostatic force on an arbitrarily inclined surface¶

In order to determine the force acting on a vertical place surface, as in the case of the swimming pool, it is better to generalise the problem and consider any surface at some angle on inclination $\theta$. The 3D model below shows such a generalised shape. A coordinate system $(x, y, z)$ is inclined at the angle $\theta$ along the wall of interest.

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Taking a view of the inclined side wall we can take an arbitrary area A as shown:¶

If we consider an infinitesimal element of area $dA$ residing in the arbitrary area $A$ on the inclined wall, we can use what we already know compute the hydrostatic pressure $dp$ at the depth $h$.

\begin{align} dp = \rho g h = \gamma h \end{align}

The force $dF$ acting on the element is then:

\begin{align} dF = \gamma h~{d\!A} = \gamma [\mathrm{y}\sin{\theta}]~{d\!A} \end{align}

Integrating over the entirety of area $A$ allows us to compute the resultant force from the forces acting on each element of $A$.

\begin{align} F_R = \gamma \sin(\theta) \int_A \mathrm{y}~{d\!A} \end{align}

You may recognise that the integral takes the form of the First Moment of Area:

\begin{align} \int_A \mathrm{y}~{d\!A} = \mathrm{y_c} A \end{align}

Which gives us:

\begin{align} F_R = \gamma \sin(\theta)~\mathrm{y_c} A \end{align}

Where $\mathrm{y_c}$ is the position of the centroid of $A$ along the local $y$-axis.

We can write this in terms of $h_c$:

\begin{align} F_R = \gamma h_c A \end{align}

However you will notice in the diagram below that the force $F_R$ does not pass through the centroid located at a depth $h_c$. This makes intuitive sense. The lower elements in area $A$ will experience greater force $dF$ than those closer to the surface. Therefore the force will be distributed towards the bottom and the resultant of that force will therefore be lower than the centroid.

$Hydrostatic force on an inclined plane$

Next we need to compute the distance $\mathrm{y_R}$ the distance at which the force $F_R$ acts. To do this we take moments about $x$ for every force $dF$ acting at every distance $\mathrm{y}$ for every infinitesimal element in area $A$.

\begin{align} F_R \mathrm{y_R} = \int_A \mathrm{y}~{d\!F} = \gamma \sin{\theta} \int_A \mathrm{y}^2~{d\!A} \end{align}

Since $F_R = \gamma\sin{\theta}~\mathrm{y_c} A$ we can substitute and obtain:

\begin{align} \mathrm{y_R} = \frac{1}{A\mathrm{y_c}} \int_A \mathrm{y}^2~{d\!A} \end{align}

The second moment of area has the form:

\begin{align} I_x = \int_A \mathrm{y^2}~{d\!A} \end{align}

therefore:

\begin{align} \mathrm{y_R} = \frac{I_x}{A\mathrm{y_c}} \end{align}

We can employ the parallel-axis theorem to determine the second moment of area about $\mathrm{y_R}$ relative to that around the centroid.

\begin{align} I_x = I_{xc} + A\mathrm{y^2_c} \end{align}

which yields:

\begin{align} \mathrm{y_R} = \frac{I_{xc} + A\mathrm{y^2_c}}{A\mathrm{y_c}} = \frac{I_{xc}}{A\mathrm{y_c}} + \mathrm{y_c} \end{align}

We have already seen the denominator, $A\mathrm{y_c}$; it is the first moment of area and the numerator $I_{xc}$ is the second moment of area about the centroid, both of which have positive values. Therefore the distance $\frac{I_{xc}}{A\mathrm{y_c}} > 0$ which means that $\mathrm{y_R}$ is always lower than $\mathrm{y_c}$.

Combining the above figures:

Location of $F_R$ in the $x$ direction¶

To locate the point which $F_R$ acts in the $x$-direction we take moments about the $y$-axis.

\begin{align} F_R \mathrm{x_R} = \int_A \mathrm{x}~{d\!F} = \int_A \gamma \sin{\theta}~\mathrm{xy}~{d\!A} \end{align}

Since $F_R = \gamma\sin{\theta}~\mathrm{y_c} A$ we can again substitute and obtain:

\begin{align} \mathrm{x_R} = \frac{\int_A \mathrm{xy}~{d\!A}}{A\mathrm{y_c}} = \frac{I_{xy}}{A\mathrm{y_c}} \end{align}

Once more invoking the parallel-axis theorem

\begin{align} I_{xy} = I_{xyc} + A\mathrm{x_c y_c} \end{align}

we obtain:

\begin{align} \mathrm{x_R} = \frac{I_{xyc} + A\mathrm{x_cy_c}}{A\mathrm{y_c}} = \frac{I_{xyc}}{A\mathrm{y_c}} + \mathrm{x_c} \end{align}

Munsen et al.[1] provides a useful reference of a the geometric properties for a number of common shapes. A version of it is reproduced here and will also be available in the final exam.

Example¶

This example is taken from Munsen et al. [1].

A $4m$ diameter circular gate is located in the inclined wall of a large reservoir containing water. The gate is mounted on a shaft along its horizontal diameter.

For a water depth of $10m$ above the shaft determine:

the magnitude and location of the resultant force exerted on the gate by the water, and
the moment that would have to be applied to the shaft to open the gate.

$\gamma =9.8kN m^3$

Part 1¶

The resultant force, as we have seen above, may be determined from

\begin{equation*} F_R = \gamma h_c A = (9800)(10)\frac{\pi 4^2}{4} = 1.23\times 10^6~N \end{equation*}

To determine the centre of pressure we once again follow the above derivation. We have shown:

\begin{equation*} \mathrm{y_R} = \frac{I_{xc}}{A\mathrm{y_c}} + \mathrm{y_c} \end{equation*}

We need $I_{xc}$ and $y_c$ to compute $\mathrm{y_R}$. For a circle we now know

\begin{equation*} I_{xc} = \frac{\pi R^4}{4} = \frac{\pi 2^4}{4} = 12.57m^4 \end{equation*}

Next we can calculate the distance along the angled wall

\begin{equation*} y_c = \frac{h_c}{\sin{\theta}} = \frac{10}{\sin{60^\circ}} = 11.55m \end{equation*}

Now can can solve part 1.

\begin{equation*} \mathrm{y_R} = \frac{I_{xc}}{A\mathrm{y_c}} + \mathrm{y_c} = \frac{12.57}{(12.57)(11.55)} + 11.55 = 0.0866+11.55 = 11.64m \end{equation*}

Part 2¶

The moment $M$ required to open the gate may be determined by taking moments about the gate shaft. The distance $(\mathrm{y_R} − y_c )$ is where the force $F_R$ acts relative to the shaft.

\begin{align*} M &= F_R (\mathrm{y_R} − y_c )\\ &= (1.23\times 10^6)(0.0866 ) = 1.07\times 10^5 Nm \end{align*}

Introduction¶

Learning outcomes¶

Hydrostatic force on plane surfaces¶

Hydrostatic force on an arbitrarily inclined surface¶

Taking a view of the inclined side wall we can take an arbitrary area A as shown:¶

Location of $F_R$ in the $x$ direction¶

Example¶

Part 1¶

Part 2¶

The Pressure Prism¶

Hydrostatic force on a curved surface¶

Buoyancy¶

Finishing up¶

References¶